Stoichiometry Chapter 9 Unit 6 Stoichiometry Chapter 9

Stoichiometry Chapter 9

Chapter 9 – Section 1: Introduction to Stoichiometry What is Stoichiometry? Stoichiometry is the branch of chemistry that deals with the mass relationships of elements in a chemical reaction. Stoichiometry calculations always start with a balanced chemical equation.

Chapter 9 – Section 1: Introduction to Stoichiometry Mole Ratio A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al2O3(l) → 4Al(s) + 3O2(g) Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al 4 mol Al 3 mol O2 3 mol O2

Chapter 9 – Section 1: Introduction to Stoichiometry Molar Mass (Review) Molar Mass is the mass of one mole of a pure substance. Molar mass units are g/mol. The molar mass of an element is the same number as its atomic mass, only the units are different. Example: Molar mass of water (H2O). (2 mol H x 1.01) + (1 mol O x 16.00) = 18.02 g/mol.

Mole to Mole Conversions Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole to Mole Conversions Using the mole ratio, you can convert from moles of one substance to moles of any other substance in a chemical reaction. Example: For the reaction N2 + 3H2 → 2NH3, how many moles of H2 are required to produce 12 moles of NH3? Use mole ratio as a conversion factor This is what we’re given: 3 mol H2 12 mol NH3 x = 18 mol H2 2 mol NH3

Mole/Mole Conversions Sample Problem 1 Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole/Mole Conversions Sample Problem 1 In a spacecraft, the CO2 exhaled by astronauts can be removed by the following reaction with LiOH: CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l) How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day? Solution: Conversion factor Given 2 mol LiOH 20 mol CO2 x = 40 mol LiOH 1 mol CO2

Mole to Mass Conversions Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole to Mass Conversions To convert from moles of one substance to grams of another, you need 2 conversion factors: 1. mole ratio. 2. molar mass of the unknown. To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.

Mole to Mass Conversions (continued) Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole to Mass Conversions (continued) Example: Given the equation: 2Mg(s) + O2(g)→ 2MgO(s), Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium. 2nd C.F. Molar Mass of the Unknown 1st C.F. Mole Ratio Given 2 mol MgO 40.3 g MgO 2.00 mol Mg x x 80.6 g MgO = 2 mol Mg 1 mol MgO

Mass to Mole Conversions Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mass to Mole Conversions To convert from grams of one substance to moles of another, you need 2 conversion factors: 1. molar mass of the given. 2. mole ratio. Mass to mole conversion factors are the inverse of mole to mass conversion factors.

Mass to Mole Conversions (continued) Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mass to Mole Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many moles of HgO are needed to produce 125g of O2? 1st C.F. Molar Mass of the Given 2nd C.F. Mole Ratio Given 1 mol O2 2 mol HgO 125 g O2 x x = 7.81 mol HgO 32 g O2 1 mol O2

Mole/Mass Conversions Sample Problem 1 Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole/Mass Conversions Sample Problem 1 The balanced equation for photosynthesis is as follows: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) What mass, in grams, of glucose (C6H12O6) is produced when 3.00 mol of water react with carbon dioxide? Solution: 2nd C.F. Molar Mass of the Unknown 1st C.F. Mole Ratio Given 1 mol C6H12O6 180.2 g C6H12O6 3.00 mol H2O x x 90.1 g C6H12O6 = 6 mol H2O 1 mol C6H12O6

Mole/Mass Conversions Sample Problem 2 Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mole/Mass Conversions Sample Problem 2 The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) If the reaction is run using 824 g NH3 and excess oxygen, how many moles of NO are formed? Solution: 1st C.F. Molar Mass of the Given 2nd C.F. Mole Ratio Given 1 mol NH3 4 mol NO 824 g NH3 x x = 48.5 mol NO 17.0 g NH3 4 mol NH3

Mass to Mass Conversions Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mass to Mass Conversions To convert from grams of one substance to grams of another, you need 3 conversion factors: 1. molar mass of the given. 2. mole ratio. 3. molar mass of the unknown.

Mass to Mass Conversions (continued) Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mass to Mass Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many grams of HgO are needed to produce 45g of O2? 3rd C.F. Molar Mass of the Unknown 1st C.F. Molar Mass of the Given 2nd C.F. Mole Ratio Given 1 mol O2 2 mol HgO 216.6 g HgO 45 g O2 x x x = 609.2 g HgO 32 g O2 1 mol O2 1 mol HgO

Mass/Mass Conversions Sample Problem 1 Chapter 9 – Section 2: Ideal Stoichiometric Calculations Mass/Mass Conversions Sample Problem 1 Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the following reaction : Sn(s) + 2HF(g) → SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.0 g HF with Sn? Solution: 3rd C.F. Molar Mass of the Unknown 1st C.F. Molar Mass of the Given 2nd C.F. Mole Ratio Given 1 mol HF 1 mol SnF2 156.7 g SnF2 117.5 g SnF2 30.0 g HF x x x = 20.0 g HF 2 mol HF 1 mol SnF2

Chapter 9 – Section 3: Limiting Reactants and Percentage Yield When combining 2 or more different things to make a product, you have to stop when one of the things is used up. For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. 

Limiting Reactants (continued) Chapter 9 – Section 3: Limiting Reactants and Percentage Yield Limiting Reactants (continued) The limiting reactant is the reactant that limits the amount of product formed. The excess reactant is the substance that is not used up completely. Once the limiting reactant is used up, a chemical reaction will stop.

Limiting Reactants (continued) Chapter 9 – Section 3: Limiting Reactants and Percentage Yield Limiting Reactants (continued) Example: Silicon dioxide reacts with hydrogen fluoride according to the following equation: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant? The limiting reactant makes the least product. Mole ratio C.F. Set up 2 equations, one for each given: 1 mol SiF4 6 mol HF 6 mol HF x = 1.5 mol SiF4 1.5 mol SiF4 4 mol HF Limiting Reactant is HF 1 mol SiF4 4.5 mol SiO2 x = 4.5 mol SiF4 1 mol SiO2

Limiting Reactants Sample Problem 1 Chapter 9 – Section 3: Limiting Reactants and Percentage Yield Limiting Reactants Sample Problem 1 According to the following reaction : CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) If 1 mol CS2 reacts with 1 mol O2, identify the limiting reactant. How many moles of excess reactant remain? x 2 mol SO2 = 2 mol SO2 1 mol CS2 Limiting Reactant is O2 1 mol CS2 1 mol O2 1 mol O2 x 2 mol SO2 = 0.67 mol SO2 0.67 mol SO2 3 mol O2 Mol of excess reactant that was used up Work backwards from actual mol of product formed 0.67 mol SO2 x 1 mol CS2 = 0.33 mol CS2 2 mol SO2 Moles remaining = 1 mol CS2 - 0.33 mol CS2 0.67 mol CS2

Chapter 9 – Section 3: Limiting Reactants and Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry). The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem). The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

Percentage Yield (continued) Chapter 9 – Section 3: Limiting Reactants and Percentage Yield Percentage Yield (continued) Example: Given the following equation: C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) When 36.8 g C6H6 react with excess Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl? This is the Theoretical Yield Set up a mass to mass conversion 36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.5 g C6H5Cl = 53.1 g C6H5Cl g C6H6 78.0 1 mol C6H6 1 mol C6H5Cl Actual 38.8 g C6H5Cl Percent Yield x 100 x 100 = 73.1% Now use the formula = = Theoretical 53.1 g C6H5Cl

Percentage Yield Sample Problem 1 Chapter 9 – Section 3: Limiting Reactants and Percentage Yield Percentage Yield Sample Problem 1 According to the following reaction : CO(g) + 2H2(g) → CH3OH(l) If the typical yield is 80%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess hydrogen gas? This is the Theoretical Yield Set up a mass to mass conversion 75.0 g CO x 1 mol CO x 1 mol CH3OH x 32.0 g CH3OH = 85.7 g CH3OH 28.0 g CO 1 mol CO 1 mol CH3OH Multiply theoretical yield by percentage to get actual yield 85.7 g CH3OH x 80% 68.6 g CH3OH =